$4800 is 60% of what number?

Choose the symbol that correctly compares the fractions below.<br> 8/9 ? 1/9

gladu [14]

Answer:

Which means that this equation is also true: 8/9 > 1/9

Step-by-step explanation:

Is 8/9 less than 1/9? Is 8/9 smaller than 1/9? These are the same questions with one answer.

To get the answer, we first convert each fraction into decimal numbers. We do this by dividing the numerator by the denominator for each fraction as illustrated below:

8/9 = 0.889

1/9 = 0.111

Then, we compare the two decimal numbers to get the answer.

0.889 is not less than 0.111.

Therefore, 8/9 is not less than 1/9 and the answer to the question "Is 8/9 less than 1/9?" is no.

5 0

3 years ago

Doss [256]

Answer:

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

A study suggested that children between the ages of 6 and 11 in the US have an average weightof 74 lbs, with a standard deviation of 2.7 lbs.

This means that $\mu = 74, \sigma = 2.7$

What proportion of childrenin this age range between 70 lbs and 85 lbs.

This is the pvalue of Z when X = 85 subtracted by the pvalue of Z when X = 70. So

X = 85

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{85 - 74}{2.7}$

$Z = 4.07$

$Z = 4.07$ has a pvalue of 1

X = 70

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{70 - 74}{2.7}$

$Z = -1.48$

$Z = -1.48$ has a pvalue of 0.0694

1 - 0.0694 = 0.9306

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

7 0

2 years ago

Which of the fractions shown in the box below is<br> the largest?<br> 3/4, 5/6, 11/12, 19/24

ruslelena [56]

Answer:

24 is the largest fraction fell by 12 to 19

6 0

3 years ago

Help on my quiz i’m so confused

DerKrebs [107]

Answer:

A

Step-by-step explanation:

6 0

3 years ago

The speeds of vehicles traveling on a highway are normally distributed with an unkown population mean and standard deviation. A

Maksim231197 [3]

Answer:

The 90% confidence interval would be given by (60.09;69.91)

We are 90% confident that the true mean for the speeds of vehicles traveling on a highway is between 60.09 and 69.91 miles per hour.

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =65$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=9 represent the sample standard deviation

n=11 represent the sample size

2) Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=11-1=10$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,10)".And we see that $t_{\alpha/2}=1.81$

Now we have everything in order to replace into formula (1):

$65-1.81\frac{9}{\sqrt{11}}=60.09$

$65+1.81\frac{9}{\sqrt{11}}=69.91$

So on this case the 90% confidence interval would be given by (60.09;69.91)

We are 90% confident that the true mean for the speeds of vehicles traveling on a highway is between 60.09 and 69.91 miles per hour.

3 0

3 years ago